User:Jesushaces/Proof

This article presents background and proofs of the fact that the recurring decimal 0.9999… equals 1, not approximately but exactly, as well as some of the most common arguments that claim that 0.999… is less than 1, or that it is not exactly 1.

Arguments that claim that 0.999… is not exactly 1

Should we include a counter-example for (each of) these claims? or an example?

Possibly the simplest argument offered as to why $0.999\ldots \neq 1$ works along the lines of "because it starts with zero, $0.999\ldots$ is obviously less than one". However, as many mathematics and science teachers are keen to point out, that which is "obvious" is not necessarily true, and such a statement is not particularly useful unless backed with some evidence that it can be proven.

The argument may be strengthened by a suggestion along the following lines:

0.9 < 1

0.99 < 1

0.999 < 1

and following this pattern, $0.999\ldots$ < 1. The problem with this argument, however, is that it suggests that a statement S(n) that "a zero followed by a decimal point followed by n 9s represents a number less than 1", which is true for all integer n is also true when n is an infinite value. The problem here is that such a deduction is not generally valid (compare with a statement such as "n is finite", which by definition is true for any integer n but false if n is infinite). In fact, the behaviour of infinite properties when compared to their finite counterparts means it is actually possible to prove statements such as $0.999\ldots \geq 1$, and with greater rigor, in some cases, than it is to prove $0.999\ldots \leq 1$.

• Infinite decimals are approximations.

• This argument claims that while it is possible to write an infinite decimal expansion such as $0.999\ldots$, this is only an approximation to the exact value and therefore cannot equal 1.

• Limits are approximations.
• Decimals and Limits are processes, not numbers.
• Infinite decimals and limits are the result of an infinite process.

• This may be considered a relation to the constructive school of mathematics, in that both the full expression of $0.999\ldots$ and the limit of partial sums that equals 1 both require an infinite number of steps, and thus cannot be "truly" calculated.

• An infinite sum is not the same as the limit of an infinite sum.

• This is an attempt to say that while $0.999\ldots =\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}$ and $\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}=1$, the infinite sum and the infinite limit of the partial sums are not the same (despite such a statement being a common definition to give the infinite sum a valid meaning) and so the numbers are equal. However, it is possible with a little effort to show that the two are in fact equal, if a few simple definitions are agreed upon.

See also:

• Talk:Proof_that_0.999..._equals_1/Archive02#If_I_may_speak_to_the_article_itself...

References:

• Conflicts in the Learning of Real Numbers and Limitsby D. O. Tall & R. L. E. Schwarzenberger, University of Warwick Published in Mathematics Teaching, 82, 44–49 (1978). pdf webpage

See also

• Recurring decimal
• Geometric series
• Convergent series
• Infinite series
• Limit (mathematics)

External links

• Why does 0.9999... = 1 ?
• Ask A Scientist: Repeating Decimals
• Repeating Nines
• Is 0.999...=1?